Tension in a Simple Pendulum: Calculating Tension at Rest and at an Angle

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SUMMARY

The tension in a simple pendulum with a 100g bob and a 1-meter string is calculated to be 0.98N when at rest (0°) and approximately 0.15N at a 30° angle from the vertical. The calculations utilize the formula F_t = -mgcos(Θ), where g is -9.8 m/s². It is important to note that when the pendulum exceeds a 15° swing angle, it no longer adheres to the simple pendulum model, complicating the physics involved. Additionally, the tension at any angle requires knowledge of the pendulum's velocity at that point for accurate determination.

PREREQUISITES
  • Understanding of basic physics concepts, specifically forces and tension.
  • Familiarity with trigonometric functions, particularly cosine.
  • Knowledge of pendulum dynamics and the limitations of the simple pendulum model.
  • Ability to perform calculations involving mass, gravity, and angles.
NEXT STEPS
  • Study the effects of velocity on tension in pendulum systems.
  • Learn about the limitations of the simple pendulum approximation and when to apply more complex models.
  • Explore the derivation of the pendulum period using differential equations.
  • Investigate centripetal acceleration and its impact on forces in swinging pendulums.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and pendulum motion, as well as educators seeking to clarify concepts related to tension and forces in pendulum systems.

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Homework Statement



A 100g pendulum bob hangs from a 1 meter string. What is the tension in the string when the pendulum is not moving (i.e. at the 'bottom')? What is the tension when it is at 30° to the vertical?

Homework Equations



100g=.1kg
g=-9.8m/s^2
F_t=-mgcos(Θ)

The Attempt at a Solution



Tension when not moving:
F_t=-mgcos(Θ)=-(.1kg)(-9.8m/s^2)cos(0)=.98N

Tension at 30° to the vertical:
F_t=-mgcos(Θ)=-(.1kg)(-9.8m/s^2)cos(30°)≈.15N

Is that correct?
My main dilemma is that my professor told us when a pendulum exceeds a 15° swing angle, that it can no longer be considered a simple pendulum and goes by more complicated physics, however, this pendulum swings to twice that angle.
Also, when I put this on a different site, someone answered and said "The tension at any angle cannot be known unless you also know its velocity at that point."

Thoughts?
 
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gmmstr827 said:

Homework Statement



A 100g pendulum bob hangs from a 1 meter string. What is the tension in the string when the pendulum is not moving (i.e. at the 'bottom')? What is the tension when it is at 30° to the vertical?

Homework Equations



100g=.1kg
g=-9.8m/s^2
F_t=-mgcos(Θ)

The Attempt at a Solution



Tension when not moving:
F_t=-mgcos(Θ)=-(.1kg)(-9.8m/s^2)cos(0)=.98N

Tension at 30° to the vertical:
F_t=-mgcos(Θ)=-(.1kg)(-9.8m/s^2)cos(30°)≈.15N

Is that correct?
My main dilemma is that my professor told us when a pendulum exceeds a 15° swing angle, that it can no longer be considered a simple pendulum and goes by more complicated physics, however, this pendulum swings to twice that angle.
Also, when I put this on a different site, someone answered and said "The tension at any angle cannot be known unless you also know its velocity at that point."

Thoughts?

Your professor is correct in terms of the simple pendulum approximation for period. In solving the differential equation to come up with T=2*Pi*sqrt(L/g), the assumption was made that sin(Θ)~~Θ, but really, that truncates the series representation of sin(Θ) at only one term. To be accurate, one should go out more terms, but it gets rather complicated quickly.

If the pendulum is swinging through the bottom, it is experiencing a greater tension because it also has a net acceleration (T exceeds mg, giving the centripetal acceleration up to the point of rotation), and by extension, it has a different tension at all points on the path of its swing if it has a non-zero velocity, changing velocity due to the centripetal acceleration.

The question would be clearer if it asked "What is the tension in the string when it is held stationary at 30deg from the vertical, by a force acting perpendicular to the vertical (i.e. your hand)." Otherwise, you have an unbalanced force and it makes for a bit of confusion.
 

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